An integer-valued function $f$ is called tenuous if $f(x) + f(y) > y^2$ for all positive integers $x$ and $y.$  Let $g$ be a tenuous function such that $g(1) + g(2) + \dots + g(20)$ is as small as possible.  Compute the minimum possible value for $g(14).$
Solution: Let $S = g(1) + g(2) + \dots + g(20).$  Then by definition of a tenuous function,
\begin{align*}
S &= [g(20) + g(1)] + [g(19) + g(2)] + [g(18) + g(3)] + \dots + [g(11) + g(10)] \\
&\ge (20^2 + 1) + (19^2 + 1) + (18^2 + 1) + \dots + (11^2 + 1) \\
&= 2495
\end{align*}Let's assume that $S = 2495,$ and try to find a function $g(x)$ that works.  Then we must have
\begin{align*}
g(20) + g(1) &= 20^2 + 1, \\
g(19) + g(2) &= 19^2 + 1, \\
g(18) + g(3) &= 18^2 + 1, \\
&\dots, \\
g(11) + g(10) &= 11^2 + 1.
\end{align*}If $g(1) < g(2),$ then
\[g(19) + g(1) < g(19) + g(2) = 19^2 + 1,\]contradicting the fact that $g$ is tenuous.

And if $g(1) > g(2),$ then
\[g(20) + g(2) < g(20) + g(1) = 20^2 + 1,\]again contradicting the fact that $g$ is tenuous.  Therefore, we must have $g(1) = g(2).$

In the same way, we can prove that $g(1) = g(3),$ $g(1) = g(4),$ and so on, up to $g(1) = g(10).$  Hence,
\[g(1) = g(2) = \dots = g(10).\]Let $a = g(1) = g(2) = \dots = g(10).$  Then $g(n) = n^2 + 1 - a$ for all $n \ge 11.$  Since $g(11) + g(11) \ge 122,$ $g(11) \ge 61.$  But $g(11) = 121 + 1 - a = 122 - a \le 61,$ so $a \le 61.$  The smallest possible value of $g(14)$ is then $14^2 + 1 - 61 = \boxed{136}.$